1. Environmental and Exploration Geophysics I tom.h.wilson tom.wilson@geo.wvu.edu Department of Geology and Geography West Virginia University Morgantown, WV Terrain Conductivity Phone - 293-6431

3. The terrain conductivity method employs an alternating electromagnetic source to induce current flow in subsurface materials. The combined source or primary field and induced or secondary field is measured by a passive receiver coil. Recall that the flow of current creates a magnetic field. Current flowing in a wire as shown at right will generate a magnetic field that encircles the wire. The magnetic force is everywhere tangential to the wire and decreases in intensity with distance from the wire Remember the right hand rule!

4. This relationship is known as the Lorentz relation. Terms like With the bar across the top indicate that we are considering the orientation of this quantity along with its amplitude This is a type of vector multiplication referred to as the cross product

5. Geometrically, the cross product is defined as shown below If v and B are perpendicular the result is just the product vB; however, if there is some angle  between these two vectors, then the result becomes vBsin .

6. Now consider the powerful magnet portrayed at the right. Its poles have been bent around to face each other. The field lines exit one pole (+) and enter the other (-). The drawing illustrates the basic rules for depicting field line orientations and polarity. Imagine that you throw an electron into the field between the two poles of the magnet. What will happen to it? The behavior of the electron in the magnetic field is defined by another “right hand” rule.

7. Connect the tale of electron velocity vector to the tale of the magnetic field vector The particle will turn and circle about the field lines. The sign of the charge is important!

8. The magnetic field lines are pointing into the board The electron moves along a clockwise curving path

9. From Halliday and Resnick What is the orientation of B? B points toward you.

10. Recall that the effect of current flow through a coil is to produce a magnetic field like that of a bar magnet.

11. You may find it useful to review basic relationships associated with the interaction of electric and magnetic fields. Recall Faraday’s Law of Induction - - and Lenz’s law, which states that “the induced current will appear in such a direction that it opposes the change that produced it.”  induced electromotive force  change in magnetic flux  t change of time

12. Movement of a magnetic field through a coil produces a change of flux enclosed by the coil of wire, which leads to current flow wihtin the coil. Faraday’s Law

13. Illustration of Lenz’sLaw The field produced by the induced current opposes the field of the magnetic field introduced into the loop

15. The terrain conductivity method offers certain benefits over the alternative electrical resistivity method. The terrain conductivity meter evaluates subsurface electrical properties without the necessity of making direct contact with the ground surface. Its effectiveness is not limited by the presence of highly resistive layers in the near surface. The terrain conductivity survey can be conducted much more rapidly. BENEFITS LIMITATIONS It is difficult to collect extensive sounding-type data because of the limited range of intercoil spacings available on conventional terrain conductivity meters.

16. We begin by covering some basic definitions Ohm’s Law V is potential difference i current R resistance Resistance, defined as opposition to direct current flow, is not a fundamental physical attribute of materials since it varies depending on the conductor geometry. The geometrical influences are evident in this relationship where  is the resistivity, l the conductor length, and A the cross-sectional area of the conductor

17. The resistivity  represents a fundamental physical property of the conductor, and this or its inverse  (the conductivity) are the parameters we wish to measure. In general - Resistivity is the property of a material which resists current flow. A l  or 

18. Units - The unit of resistance is the ohm Balancing units in the definitional formula for resistivity, we see that resistivity has units of ohm-meters or  -m. Conductance is the reciprocal of resistance and has units of ohm -1. Thus conductivity (  ) has units of ohm -1 /m or mho/meter

19. The reciprocal of a resistivity of 1  -m corresponds to a conductivity of 1 mho/meter 1 mho/meter = 1000 millimhos/meter 1 millimho/meter =0.001 mho/meters The reciprocal of 0.001 mho/meters is 1000  -m Working back and forth between units of conductivity and resistivity Conductance (1/R) is often measured in units of Seimens (S) which are equivalent to mhos. 1 S = 1 mho

20. In general when given a resistance the equivalent conductivity in millimhos/meter is obtained by taking the inverse of the resistivity and multiplying by 1000. The same applies to the computation of resistivity when given the conductivity. 100  -m = _____ millimhos/meter 20 millimhos/meter = _____  -m 10 50

21. Factors Affecting Terrain Conductivity 1. Porosity: shape and size of pores, number 2. Permeability: size and shape of interconnecting passages 3. The extent to which pores are filled by water, i.e. the moisture content 4. Concentration of dissolved electrolytes 5. Temperature and phase state of the pore water 6. Amount and composition of colloids

23. Clay particles are a source of loosely held cations

24. Cation clouds provide a source of electrolytes, they can also form a partial barrier to current flow through small pores. In this case their effect is similar to that of a capacitor.

25. Electromagnetic terrain conductivity measurements at low induction numbers J. D. McNeill, Technical Note TN-6: Geonics LTD.

26. Note that the relative response function for the horizontal dipole  h is much more sensitive to near-surface conductivity variations and that its response or sensitivity drops off rapidly with depth Vertical dipole interaction has no sensitivity to surface conductivity, reaches peak sensitivity at z ~0.5, and is more sensitive to conductivity at greater depths than is the horizontal dipole.

27. The contribution of this thin layer to the overall ground conductivity is proportional to the value of the relative response function at that depth. Constant conductivity 

28. The contribution of a layer to the overall ground conductivity is proportional to the area under the relative response function over the range of depths (Z 2 -Z 1 ) spanned by that layer. Z1Z1 Z2Z2

29. As you might expect, the contribution to ground conductivity of a layer of constant conductivity that extends significant distances beneath the surface (i.e. homogenous half-space) is proportional to the total area under the relative response function.

30. The terrain conductivity meter gives us a single number –  a – the apparent conductivity So in general the contribution of each layer to the overall ground conductivity – the apparent conductivity - will be in proportion to the areas under the relative response function spanned by each layer.  a = f(  1,  2,  3 ) 11 22 33

31. You all will recognize these area diagrams as integrals. The contribution of a given layer to the overall ground conductivity at the surface above it is proportional to the integral of the relative response function over the range of depths spanned by the layer.

32. McNeill introduces another function, R(z) - the cumulative response function - which he uses to compute the ground conductivity from a given distribution of conductivity layers beneath the surface. For next time continue your reading of McNeill and develop a general appreciation of the relative and cumulative response functions.

33. Each point on the R V (z) curve represents the area under the  V (z) curve from z to . The following diagrams are intended to help you visualize the relationship between R(z) and  (z).

34. Z2Z2

37. Consider one additional integral - How would you express this integral as a difference of cumulative response functions?

39. Note that R(0) = 1, hence

40. According to our earlier reasoning - the contribution of a single conductivity layer to the measured ground (or terrain) conductivity is proportional to the area under the relative response function. The apparent conductivity measured by the terrain conductivity meter at the surface is the sum total of the contributions from all layers. We know that each of the areas under the relative response curve can be expressed as a difference between cumulative response functions

41. Let’s consider the following problem, which is taken directly from McNeill’s technical report (TN6). See also Box 11.2 and Figure 11.7 (p 594 & 596 of Reynolds) Error in Burger, Sheehan and Jones

44. Visually, the solution looks like this..

45. Compare this result to that of McNeill’s (see page 8 TN6). Mathematical formulation -

46. In this relationship  a is the apparent conductivity measured by the conductivity meter. The dependence of the apparent conductivity on intercoil spacing is imbedded in the values of z. Z for a 10 meter intercoil spacing will be different from z for the 20 meter intercoil spacing. The above equation is written in general form and applies to either the horizontal or vertical dipole configuration.

47. In the appendix of McNeill (today’s handout), he notes that the assumption of low induction number yields simple algebraic expressions for the relative and cumulative response functions. We can use these relationships to compute specific values of R for given zs.

48. The simple algebraic expressions for R V (z) and R H (z) make it easy for us to compute the terms in the problem McNeill gives us. In that problem z 1 is given as 0.5 and z 2 as 1 and 1.5 Assuming a vertical dipole orientation R V (z=0.5) ~ 0.71 R V (z=1.0) ~ 0.45 R V (z=1.5) ~ 0.32

49.  1 =20 mmhos/m  2 =2 mmhos/m  3 =20 mmhos/m Z 1 = 0.5 Z 2 = 1 and 1.5 Substituting in the following for the case where Z 2 =1.

50. Start looking over problems 8.5, 8.6, and 8.7 AMD problem will be due next Tuesday (bring questions to class on Thursday) Bring questions for discussion to class on Tuesday. Papers on the Terrain Conductivity reading list are available in the 3 rd floor mail room.

52. Here’s the problem handed out to you last Thursday A terrain conductivity survey is planned using the EM31 meter (3.66 m (or 12 foot) intercoil spacing). Our hypothetical survey was conducted over a mine spoil to locate migration pathways within the spoil through which acidic mine drainage as well as neutralizing treatment are being transported. Scattered borehole data across the spoil suggest that these paths are approximately 10 feet thick and several meters in width. Borehole resistivity logs indicate that areas of the spoil surrounding these conduits have low conductivity averaging about 4mmhos/m. The bedrock or pavement at the base of the spoil also has a conductivity of approximately 4mmhos/m. Depth to the pavement in the area of the proposed survey is approximately 60 feet. Conductivity of the AMD transport channels is estimated to be approximately 100mmhos/m. Mine spoil surface AMD contamination zone Pit Floor ~10ft ~60ft

53. A. Evaluate the possibility that the EM31 will be able to detect high conductivity transport zones with depth-to-top of 30feet. Evaluate only for the vertical dipole mode. It may help to draw a cross section.  s = 100 mmhos/m  b = 4 mmhos/m  br = 4 mmhos/m

54. How many different conductivity layers will you actually have to consider? Does it matter whether d (depth) and s (intercoil spacing) are in feet or meters? Set up your equation following the example presented by McNeill and reviewed in class. Solve for the apparent conductivity recorded by the EM31 over this area of the spoil. Note - a table of R values are presented on the following page.  s = 100 mmhos/m  b = 4 mmhos/m  br = 4 mmhos/m

55. Z R V R H.000 1.000000 1.000000.200.9284767.6770329.400.7808688.4806249.600.6401844.3620499.800.5299989.2867962 1.000.4472136.2360680 1.200.3846154.2000000 1.400.3363364.1732137 1.600.2982750.1526108 1.800.2676438.1363084 2.000.2425356.1231055 2.200.2216211.1122055 2.400.2039542.1030602 2.600.1888474.0952811 2.800.1757906.0885849 3.000.1643990.0827627 3.200.1543768.0776539 3.400.1454940.0731363 3.600.1375683.0691128 3.800.1304545.0655074 4.000.1240347.0622578 4.200.1182129.0593147 4.400.1129097.0566359 4.600.1080592.0541887 4.800.1036061.0519428 5.000.0995037.0498762 5.200.0957124.0479660 5.400.0921982.0461979 5.600.0889320.0445547 5.800.0858884.0430231  s = 100 mmhos/m  b = 4 mmhos/m  br = 4 mmhos/m Cumulative Response Functions